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lutik1710 [3]
3 years ago
6

An alternating current E(t) =120 sin(12t) has been running through a simple circuit for a long time. The circuit has an inductan

ce of L = 0.2 henrys, a resistor of R = 5 ohms and a capacitor of capcitance C = 0.043 farads. What is the amplitude of the current I?
Engineering
1 answer:
german3 years ago
4 0

Answer:

Explanation:

we have given E(t)=120 sin(12t)

R=5 ohm

L=0.2 H

ω=12 ( from expression of E)

X_L=0.2\times 12=2.4 ohm

X_C=\frac{1}{\omega \times C}=\frac{1}{12\times 0.043}=1.9379\ ohm

Z=\sqrt{R^2+\left ( \omega L-\frac{1}{\omega C} \right )^2}

Z=\sqrt{5^2+\left ( \2.4-1.9379 )^2}

=5.021 ohm

so amplitude of current =  \frac{v}{z}=\frac{120}{5.021}=23.89

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A) P(W) = 0.5

B) P(MF) = 0.3

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A) Now, phones can either be fast(F) or slow(W). Thus, we can write;

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We are given P[F] = 0.5 and P[HF] = 0.2.

P(H ∩ F) is same as P[HF]

Also, P(M ∩ F) is same as P(MF)

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P(M) = 0.1 + 0.3

P(M) = 0.4

Now, since cellular phones can either be handheld(H) or Mobile(M), then we can say;

P(H) + P(M) = 1

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